Week 8 - Support Vector Machines

Dr. David Elliott

1.1. Introduction

1.2. What is a Hyperplane?

1.3. Maximal Margin Classifier

TODO

• look through https://github.com/JWarmenhoven/ISLR-python/blob/master/Notebooks/Chapter%209.ipynb, and https://scikit-learn.org/stable/auto_examples/exercises/plot_iris_exercise.html#sphx-glr-auto-examples-exercises-plot-iris-exercise-py
• Tidy up the code... I dont need to worry about code blocks as much now as I don't need to show any code on a slide now!

1.1. Introduction

The term Support Vector Machines (SVM's) is sometimes used loosely to refer to three methods, each an extension of the previous method¹:

• Maximal margin classifier,
• Support vector classifier,
• Support vector machine.

SVM's are a common supervised discriminative algorithm, well suited to complex small- to medium sized datasets²

They can be used for both classification and regression.

Dataset Example: Iris

To demonstrate SVM's I'll be using Fisher's (or Anderson's) Iris flowers dataset³.

The dataset consists of 50 samples each from three species of Iris (Iris setosa, Iris virginica and Iris versicolor).

Notes

• "...This data comes from a famous experiment on a series of measurements of three species of iris flowers. R A Fisher, a statistically minded thinker in the early twentieth centure used this dataset in his 1936 paper The Use of multiple measurements in taxonomic problems, published in the Annals of Eugenics."⁹

Figure 1: Iris Flowers

Five attributes were collected for the 150 records.

species (the target)

• sepal length (cm)
• sepal width (cm)
• petal length (cm)
• petal width (cm)
• species (the target)

	Sepal length (cm)	Sepal width (cm)	Petal length (cm)	Petal width (cm)	Species
0	5.1	3.5	1.4	0.2	Setosa
1	4.9	3.0	1.4	0.2	Setosa
2	4.7	3.2	1.3	0.2	Setosa
3	4.6	3.1	1.5	0.2	Setosa
4	5.0	3.6	1.4	0.2	Setosa

Notes

• Species is the typical target for its use in classification

Figure 2: Iris Attributes

Notes

• We could use any one of the decision boundries above to separate the flower classes, but which one if the "best"?

Notes

• The middle thick line is a hyperplane and the dashed outer lines are the edges of the margin.
• The circled points are the class examples that fall on the margin. These will later be termed support vectors.

1.3. What is a Hyperplane?

In $P$-dimensional space, a hyperplane is a flat affine subspace of dimension $P-1$.

Examples

• Two-dimensions: A flat one-dimensional line
• Three-dimensions: A two-dimensional subspace
• P-dimensions: A $P-1$ dimensional subspace

Notes

• "Affine" just means "does not need to pass through the origin"

• Figure Examples

  • Two-dimensions: Figures 5 & 6
  • Three-dimension: Figure 7
  • P-dimensions: ...you'll have to use your imagination.

• If you want to interact with the 3D plot use %matplotlib notebook

TODO

• Make the below function more interactive and add a title

In more detail¹:

Two-dimensions: $\beta_0 + \beta_1X_1 + \beta_2X_2 = 0$

Three-dimensions: $\beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_3 = 0$

P-dimensions: $\beta_0 + \beta_1X_1 + \beta_2X_2 + ... + \beta_pX_p = 0$

If $X = (X_1, ..., X_p)^T$ satisfies above, then it is a point on the hyperplane.

If $\beta_0 + \beta_1X_1 + \beta_2X_2 + ... + \beta_pX_p > 0$, it lies on one side of the hyperplane,

so if, $\beta_0 + \beta_1X_1 + \beta_2X_2 + ... + \beta_pX_p < 0$, its on the other side.

Notes

• In-other-words, $p$-dimensional space is dividided into two halves.

Classifying data

We aim to classify an $n \times p$ matrix of $n$ observations in $p$ dimensional space, with these observations falling into two classes $y_1,...,y_n \in \{-1,1\}$.

If we were to perfectly separate the classes the hyperplane would have the property that¹:

$\beta_0 + \beta_1X_{i1} + \beta_2X_{i2} + ... + \beta_pX_{ip} > 0 \quad \text{if} \ y_i = 1$,

$\beta_0 + \beta_1X_{i1} + \beta_2X_{i2} + ... + \beta_pX_{ip} < 0 \quad \text{if} \ y_i = -1$.

For a new test observations $x^*$, we would look at the sign of:

$$f(x^*) = \beta_0 + \beta_1x_1^* + \beta_2x_2^* + ... + \beta_px_p^*.$$

We would assign it to class 1 if $f(x^*)$ is positive and class -1 if negative.

Furthermore, we could use the magnitude of $f(x^*)$ to indicate how far the point lies from the hyperplane.

1.4. Maximal Margin Classifier

We need a reasonable way of constucting a hyperplane, out of the possible choices.

Maximal margin hyperplanes look at getting the hyperplane that is the furthest from the training observations - we compute the perpendicular distance from each training observation to a given separating hyperplane.

The maximal margin hyperplane is the separating hyperplane for which the margin is largest.

We hope the classifier with a large margin on the training data will generalise well to unseen test observations.

Another way of defining our hyperplane is:

$$\mathbf{w}^T\mathbf{x} + b = 0.$$

Our margins, where our labels $y \in \{-1,1\}$, are then

$$\mathbf{w}^T\mathbf{x} + b \geq 0 \text{ for } y_i +1$$$$\mathbf{w}^T\mathbf{x} + b < 0 \text{ for } y_i -1$$

Notes

• $\mathbf{w}$ is a weight vector
• $\mathbf{x}$ is an input vector
• $b$ is our bias

We can see below there are two equidistant points from the maximal margin hyperplane, lying on the dashed lines. These observations are called Support Vectors.

We can call these two points, $x_1$ and $x_2$ as below:

$$\mathbf{w}^Tx_1 + b = 1,$$$$\mathbf{w}^Tx_2 + b = -1.$$

If we move our support vectors our hyperplane will too.

This is because the maximal margin hyperplane only depends on these support vectors.

Other data points could be moved without the hyperplane moving.

We want to maximise the distance between the margin lines, on which the points lie⁴.

$$x_1-x_2$$

$$ \frac{\phantom{\quad}\mathbf{w}^Tx_1 + b = 1\\ - \mathbf{w}^Tx_2 + b = -1 \\}{\\ \mathbf{w}^T(x_1 - x_2) = 2} $$

$$ \frac{\mathbf{w}^T}{||\mathbf{w}||}(x_1 - x_2) = \frac{2}{||\mathbf{w}||} $$

Notes

1. we want the distance between $x_1$ and $x_2$
2. our $b$ cancels out and we end up with equation 3
3. we want to remove $\mathbf{w}^T$
4. we do this with norm of $\mathbf{w}^T$. However we cannot remove $||\mathbf{w}||$, so we are left to maximise $\frac{2}{||\mathbf{w}||}$.

$||\mathbf{w}||$ is the Euclidean norm of $\mathbf{w}$

$\text{max}\frac{2}{||\mathbf{w}||}$, while classifying everything correctly, $y_i(\mathbf{w}^T\mathbf{x}_i+b) \geq 1 \quad \forall_i$.

As our two margin lines are parallel and no training points fall between them, we can find the pair of hyperplanes which gives the maximum margin by minimizing $||\mathbf{w}||^2$. So now we have⁵:

${\text{min} \atop \mathbf{w}, b}\frac{1}{2}||\mathbf{w}||^2 \quad \text{s.t.} \quad y_i(\mathbf{w}^T\mathbf{x}_i+b) \geq 1 \quad \forall_i$,

which is easier because this is a convex quadratic optimisation problem which is efficiently solvable using quadratic programming algorithms.

Notes

• $\frac{1}{2}$ is added for convience.

Primal Problem

This requires a Lagrangian formulation of the problem so we introduce Lagrange multipliers, $\alpha_i, i = 1, \ldots , l$, where $l$ is the number of training points:

$$ \min L_P = \frac{1}{2}||\mathbf{w}||^2 - \sum^l_{i=1} \alpha_iy_i(\mathbf{x}_i\cdot\mathbf{w}+b) + \sum^l_{i=1} \alpha_i \qquad \text{s.t.} \quad \forall_i \alpha_i \geq 0 $$

Duel Problem^6,7

Removes the dependence on $\mathbf{w}$ and $b$,

$$ \max L_D = \sum_i\alpha_i - \frac{1}{2}\sum_{i,j}\alpha_i\alpha_jy_iy_j\mathbf{x}_i\cdot \mathbf{x}_j \qquad \text{s.t.} \quad \forall_i \alpha_i \geq 0. $$

To achive this we first need to set a constaint that allows us to solve for $\alpha_i$,

$$\sum_i\alpha_iy_i = 0.$$

Knowing our $\alpha_i$ means we can find the weights, which are a linear combination of the training inputs, $\mathbf{x}_i$, training outputs, $y_i$, and the values of $\alpha$,

$$\mathbf{w} = \sum^{N_S}_{i=1}\alpha_iy_i\mathbf{x}_i,$$

where $N_S$ is the number of support vectors⁶.

$b$ is then implicitly determined by choosing any $i$ for which $\alpha \neq 0$ and computing $b$ using the following Karush-Kuhn-Tucker (KKT) condition: $$ \alpha_i(y_i(\mathbf{w} \cdot \mathbf{x}_i + b) - 1) = 0 \quad \forall i $$

Notes

• $\sum_i\alpha_iy_i = 0$ removes our $b$
• There is a Lagrange multiplier $\alpha_i$ for every training point. Points for which $\alpha_i > 0$ are called "support vectors", and lie on one of the margins, with all other training points having $\alpha_i = 0$.
• Notice this means we only need to focus on those with $\alpha_i > 0$.
• $\alpha_i$ will only be zero if all training points have the same class
• Solving the SVM problem is equivalent to finding a solution to the Karush-Kuhn-Tucker (KKT) conditions.
• it is numerically safer to take the mean value of $b$ resulting from all such equations.
• For more reading on quadratic programming for support vector machines, I recommend you read:
  • Burges, C. J. (1998). A tutorial on support vector machines for pattern recognition. Data mining and knowledge discovery, 2(2), 121-167.
  • https://www.adeveloperdiary.com/data-science/machine-learning/support-vector-machines-for-beginners-duality-problem/

Inner products, similarity, and SVMs⁸

Question: But why bother doing this? That was a lot of effort, why not just solve the original problem?

Answer: Because this will let us solve the problem by computing the just the inner products of $\mathbf{x}_i$, $\mathbf{x}_j$ which will be important when we want to solve non-linearly separable classification problems.

• Inner products provide a measure of "similarity"
• Inner product in 2D between 2 vectors of unit length returns the cosine of the angle between them. In otherwords, how "far apart" they are.
  • if they are parallel their inner product is 1 (completely similar), $\mathbf{x}^T\mathbf{y} = \mathbf{x} \cdot \mathbf{y} = 1$.
  • If they are perpendicular (completely unlike) their inner product is 0 (so should not contribute to the correct classifier), $\mathbf{x}^T\mathbf{y} = \mathbf{x} \cdot \mathbf{y} = 0$.

$$ L_D(\alpha_i) = \sum_i^l\alpha_i - \frac{1}{2}\sum_{i,j}^l\alpha_i\alpha_jy_iy_j\mathbf{x}_i\cdot \mathbf{x}_j \qquad \text{s.t.} \quad \forall_i \alpha_i \geq 0, \ \sum_i^l\alpha_iy_i = 0. $$

Case 1 Two features $x_i$, $x_j$ are completely dissimilar (orthogonal), so their dot product is 0. They don't contribute to $L$.

Case 2 Two features $x_i$, $x_j$ are similar and predict the same output value $y_i$ (ie. both $+1$ or $-1$). This means $y_i \times y_j = 1$ and the value $\alpha_i\alpha_jy_iy_j\mathbf{x}_i\mathbf{x}_j$ is positive. However this decreases the value of $L$, due to subtracting from the first term sum, $\sum_i^l\alpha_i$, so the algorithm downgrades similar feature vectors that make the same prediction.

Case 3 : Two features $x_i$, $x_j$ predict opposite predictions about the output value $y_i$ (ie. one $+1$ and the other $-1$), but are otherwise similar. The value $\alpha_i\alpha_jy_iy_j\mathbf{x}_i\mathbf{x}_j$ is negative and since we are subtracting it, this adds to the sum. These are the examples we are looking for as they are critical for telling the two classes apart.

Test Phase⁶

One we have a trained Support Vector Machine, any new test data has its class determined by which side of the decision boundary the observation lands.

We take the class of $\mathbf{x^*}$ to be $\text{sgn}(\mathbf{w} \cdot \mathbf{x} + b)$.

Notes

• Remember most weights $\mathbf{w}_i$ will be 0 as only "support vectors" (on the margin/gutters) will have none 0 weights.

Limitations

Maximal Margin Classifiers are sensitive to outliers.

Notes

• In figure 12 above, we can see that the reliance on a small number of observations means there is now a small margin.
• We want to be confident that a distance from the hyperlane is a measure of our confidence in its classification, and that we have no overfit to our training data.

In other cases, no exact linear separating hyperplane exists. Therefore we may want to use a hyperplane that almost separates the two classes, allowing some errors, using a soft margin (Support Vector Classifier).

Furthermore, if we have a large number of features, this approach often leads to overfitting.

Associated Exercises

Now might be a good time to try exercises 1-3.

References

1. James, G., Witten, D., Hastie, T., & Tibshirani, R. (2013). An introduction to statistical learning (Vol. 112, p. 18). New York: springer.
2. Géron, A. (2017). Hands-on machine learning with Scikit-Learn and TensorFlow: concepts, tools, and techniques to build intelligent systems. " O'Reilly Media, Inc.".
3. Fisher, R. A. (1936). The use of multiple measurements in taxonomic problems. Annals of eugenics, 7(2), 179-188.
4. [youtube vid]
5. Murphy, K. P. (2012). Machine learning: a probabilistic perspective. MIT press.
6. Burges, C. J. (1998). A tutorial on support vector machines for pattern recognition. Data mining and knowledge discovery, 2(2), 121-167.
7. Fletcher, R. Practical Methods of Optimization. John Wiley and Sons, Inc., 2nd edition, 1987
8. http://web.mit.edu/6.034/wwwbob/svm-notes-long-08.pdf
9. https://bookdown.org/koehlerj/qr_book/introducing-the-iris-dataset.html